中五程度數學題 求解答

(a)
Locus of P:
AP^2 = 5^2 = (x + 3)^2 + (y - 4)^2
x^2 + 6x + 9 + y^2 - 8y + 16 = 25
x^2 + y^2 + 6x - 8y = 0 ... (i)

(b)
O = (0, 0)
Put O into (i)
L.H.S. = 0^2 + 0^2 + 6(0) - 8(0) = 0 = R.H.S.
i.e. O is a point on circle x^2 + y^2 + 6x - 8y = 0.

Since M is mid-point of chord OP in circle x^2 + y^2 + 6x - 8y = 0
AM ⊥ OP
i.e. ∠AMO = 90°
Therefore, A, M, O are con-cyclic with diameter AO, and
hence the center of the new circle, formed
= the mid-point of AO
= [(-3 - 0)/2, (4 - 0)/2]
= (-3/2, 2)

length of radius of new circle
= (1/2)AO
= (1/2)[(-3 - 0)^2 + (4 - 0)^2]^(1/2)
= 5/2

M is a moving point on the new circle.
i.e. Locus of M:
[x - (-3/2)]^2 + (y - 2)^2 = (5/2)^2
(x + 3/2)^2 + (y - 2)^2 = (5/2)^2
x^2 + 3x + 9/4 + y^2 - 4y + 4 = 25/4
x^2 + y^2 + 3x - 4y = 0